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Array formatting with argmax numpy Python

I am having trouble with the Numbers[(Numbers<=0).argmax():] = 0 function it is supposed to turn all the elements behind it into zeroes if the condition is met, however if the condition is not met it turns all the array elements into zeroes. How can i fix this issue. If the Numbers<=0 condition is not met the array should not change.
Array with satisfying condition at -35.15610151:
Numbers = np.array([123.6, 123.6 , 123.6, 110.3748, 111.6992976,
102.3165566, 97.81462811 , 89.50038472 , 96.48141473 , 90.49956702,
88.59907611 , 77.96718698, 61.51611052, 56.84088612, 55.36302309,
54.69866681, 56.44902415 , 59.49727145, 42.12406819, 27.42276839,
33.86711896, 32.10602877, -35.15610151, 32.34361339 , 29.20628289])
Numbers[(Numbers<=0).argmax():] = 0
Output:
[123.6 123.6 123.6 110.3748 111.6992976
102.3165566 97.81462811 89.50038472 96.48141473 90.49956702
88.59907611 77.96718698 61.51611052 56.84088612 55.36302309
54.69866681 56.44902415 59.49727145 42.12406819 27.42276839
33.86711896 32.10602877 0. 0. 0. ]
Array with no satisfying condition, turned -35.15610151 into +35.15610151:
Numbers = np.array([123.6, 123.6 , 123.6, 110.3748, 111.6992976,
102.3165566, 97.81462811 , 89.50038472 , 96.48141473 , 90.49956702,
88.59907611 , 77.96718698, 61.51611052, 56.84088612, 55.36302309,
54.69866681, 56.44902415 , 59.49727145, 42.12406819, 27.42276839,
33.86711896, 32.10602877, 35.15610151, 32.34361339 , 29.20628289])
Numbers[(Numbers<=0).argmax():] = 0
Output:
[0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0.
0.]

Try these 2 methods, one is in place assignment to the NumPy view and the other creates a fresh array to be assigned to another variable -
#Method 1 (Inplace assignment)
Numbers[(Numbers<=0).cumsum(dtype=bool)] = 0
Or,
#Method 2 (Not inplace)
np.where(~(Numbers<=0).cumsum(dtype=bool), Numbers, 0)
Or,
#As an excellent suggestion by Mad Physicist!
Numbers[np.logical_or.accumulate(Numbers >= 0)] = 0
Explanation -
The bool array that returns [F, F, F, T, F, F, F] can be seen as an array of 1s and 0s. Doing a cumsum ends up propogating the first T to the subsequent elements.
This, therefore, turns the array as [F, F, F, T, T, T, T] which can now be used with just boolean indexing and set the view to 0 OR np.where to fetch original elements or 0 based on reversing the boolean with ~
Advantage here is that if your array is just composed of False, meaning no element meets the condition, it just returns the original Numbers itself, instead of setting them to 0.
Running tests -
With a value that meets condition
Numbers = np.array([123.6 , 123.6 , -123.6, 110.3748 , 111.6992976, 102.3165566, 97.81462811])
Numbers[(Numbers<=0).cumsum(dtype=bool)] = 0
#array([123.6, 123.6, 0. , 0. , 0. , 0. , 0. ])
With no values meeting the condition
Numbers = np.array([123.6 , 123.6 , 123.6, 110.3748 , 111.6992976, 102.3165566, 97.81462811])
Numbers[(Numbers<=0).cumsum(dtype=bool)] = 0
#array([123.6 , 123.6 , 123.6 , 110.3748 ,111.6992976 , 102.3165566 , 97.81462811])
EDIT: New scenario as requested
Numbers1 = np.array([1.1, 2.2, 3.3, 4.4, 5.5])
Numbers2 = np.array([1,2,-3,4,5])
Numbers2 = np.where(~(Numbers2<=0).cumsum().astype(bool), Numbers1, 0)
Numbers2
array([1.1, 2.2, 0. , 0. , 0. ])

Just use an if for this, it represents the intention quite well, and is easy to understand:
smaller_equal_zero = Numbers <= 0
if smaller_equal_zero.any():
Numbers[smaller_equal_zero.argmax():] = 0
With Python 3.8+ you can use an assignment expression in the if:
if (smaller_equal_zero := Numbers <= 0).any():
Numbers[smaller_equal_zero.argmax():] = 0

+= with numpy.array object modifying original object

In the following code, I am attempting to calculate both the frequency and sum of a set of vectors (numpy vectors)
def calculate_means_on(the_labels, the_data):
freq = dict();
sums = dict();
means = dict();
total = 0;
for index, a_label in enumerate(the_labels):
this_data = the_data[index];
if a_label not in freq:
freq[a_label] = 1;
sums[a_label] = this_data;
else:
freq[a_label] += 1;
sums[a_label] += this_data;
Suppose the_data (a numpy 'matrix') is originally :
[[ 1. 2. 4.]
[ 1. 2. 4.]
[ 2. 1. 1.]
[ 2. 1. 1.]
[ 1. 1. 1.]]
After running the above code, the_data becomes:
[[ 3. 6. 12.]
[ 1. 2. 4.]
[ 7. 4. 4.]
[ 2. 1. 1.]
[ 1. 1. 1.]]
Why is this? I've deduced it down to the line sums[a_label] += this_data; as when i change it to sums[a_label] = sums[a_label] + this_data; it behaves as expected; i.e., the_data is not modified.

This line:
this_data = the_data[index]
takes a view, not a copy, of a row of the_data. The view is backed by the original array, and mutating the view will write through to the original array.
This line:
sums[a_label] = this_data
inserts that view into the sums dict, and this line:
sums[a_label] += this_data
mutates the original array through the view, since += requests that the operation be performed by mutation instead of by creating a new object, when the object is mutable.

Operations with columns from different files

I have many files .txt of this type:
name1.fits 0 0 4088.9 0. 1. 0. -0.909983 0.01386 0.91 0.01386 -0.286976 0.00379 2.979 0.03971 0. 0.
name2.fits 0 0 4088.9 0. 1. 0. -0.84702 0.01239 0.847 0.01239 -0.250671 0.00261 3.174 0.04749 0. 0.
#name3.fits 0 0 4088.9 0. 1. 0. -0.494718 0.01168 0.4947 0.01168 -0.185677 0.0042 2.503 0.04365 0. 0.
#name4.fits 0 1 4088.9 0. 1. 0. -0.751382 0.01342 0.7514 0.01342 -0.202141 0.00267 3.492 0.07224 0. 0.
name4.fits 0 1 4088.961 0.01147 1.000169 0. -0.813628 0.01035 0.8135 0.01035 -0.217434 0.00196 3.515 0.04045 0. 0.
I want to divide the values of one of these columns by the values of a column from another file of the same type. Here is what I have so far:
with open('4026.txt','r') as out1, open('4089.txt', 'r') as out2, \
open('4116.txt', 'r') as out3, open('4121.txt', 'r') as out4, \
open('4542.txt', 'r') as out5, open('4553.txt', 'r') as out6:
for data1 in out1.readlines():
col1 = data1.strip().split()
x = col1[9]
for data2 in out2.readlines():
col2 = data2.strip().split()
y = col2[9]
f = float(y) / float(x)
print f
However I'm getting the same values for x. For example if the first set of data is 4089.txt, and the second (4026.txt) is:
name1.fits 0 0 4026.2 0. 1. 0. -0.617924 0.01749 0.6179 0.01749 -0.19384 0.00383 2.995 0.09205 0. 0.
name2.fits 0 0 4026.2 0. 1. 0. -0.644496 0.01218 0.6445 0.01218 -0.183373 0.00291 3.302 0.05261 0. 0.
#name3.fits 0 0 4026.2 0. 1. 0. -0.507311 0.01557 0.5073 0.01557 -0.176148 0.00472 2.706 0.07341 0. 0.
#name4.fits 0 1 4026.2 0. 1. 0. -0.523856 0.01086 0.5239 0.01086 -0.173477 0.00279 2.837 0.05016 0. 0.
name4.fits 0 1 4026.229 0.0144 1.014936 0. -0.619708 0.00868 0.6106 0.00855 -0.185527 0.00189 3.138 0.04441 0. 0.
and I want to divide the 9th column of each file, taking only the first elements of each column I should get 0.91/0.6179 = 1.47, but I obtain 0.958241758242.

What's happening is that the code you have is capturing the last value in the for loop and dividing that. You should conduct the division at each stage of the for-loop to get the correct divisions.
An easier approach is placing all the values in a list
e.g.
x = [0.0149,0.01218,..etc] and y = [...]
Then you divide the two lists using numpy (or a for-loop against the lists). Remember that they both need to be of the same size to work.
Sample code:
with open('4026.txt','r') as out1, open('4089.txt', 'r') as out2, open('4116.txt', 'r') as out3, open('4121.txt', 'r') as out4, open('4542.txt', 'r') as out5, open('4553.txt', 'r') as out6:
# Build two lists
x = []
y = []
for data1 in out1.readlines():
col1 = data1.strip().split()
x.append(col1[9])
for data2 in out2.readlines():
col2 = data2.strip().split()
y.append(col2[9])
for i in range(0,len(x)):
# Make sure the denominator is not zero
if y[i] != 0:
print (1.0 * x[i])/y[i]
else:
print "Not possible"

You could do it like this:
with open('4026.txt','r') as out1, open('4089.txt', 'r') as out2:
x_col9 = [data1.strip().split()[9] for data1 in out1.readlines()]
y_col9 = [data2.strip().split()[9] for data2 in out2.readlines()]
if len(x_col9) != len(y_col9):
print('Error: files do not have same number of rows')
else:
f = [(float(y) / float(x)) for x, y in zip(x_col9, y_col9)]
print(f)
It may be better to process the files as shown below because it doesn't require reading the entire contents of all of them into memory first, and instead processes each one a line at a time:
x_col9 = [data1.strip().split()[9] for data1 in out1]
y_col9 = [data2.strip().split()[9] for data2 in out2]

Suggestions for faster for/if statements in my code?

My code takes about two hours to process. The bottleneck is in for loop and if
statements (see comment in code).
I'm beginner with python :) Can anyone recommend an efficient python way to replace the nested for and if statements?
I have tables of ~30 million rows, each row with (x,y,z) values:
20.0 11.3 7
21.0 11.3 0
22.0 11.3 3
...
My desired output is a table in the form x, y, min(z), count(min(z)). The last
column is a final count of the least z values at that (x,y). Eg:
20.0 11.3 7 7
21.0 11.3 0 10
22.0 11.3 3 1
...
There's only about 600 unique coordinates, so the output table will be 600x4.
My code:
import numpy as np
file = open('input.txt','r');
coordset = set()
data = np.zeros((600,4))*np.nan
irow = 0
ctr = 0
for row in file:
item = row.split()
x = float(item[0])
y = float(item[1])
z = float(item[2])
# build unique grid of coords
if ((x,y)) not in coordset:
data[irow][0] = x
data[irow][1] = y
data[irow][2] = z
irow = irow + 1 # grows up to 599
# lookup table of unique coords
coordset.add((x,y))
# BOTTLENECK. replace ifs? for?
for i in range(0, irow):
if data[i][0]==x and data[i][1]==y:
if z > data[i][2]:
continue
elif z==data[i][2]:
ctr = ctr + 1
data[i][3]=ctr
if z < data[i][2]:
data[i][2] = z
ctr = 1
data[i][3]=ctr
edit: For reference the approach by #Joowani computes in 1m26s. My original approach, same computer, same datafile, 106m23s.
edit2: #Ophion and #Sibster thanks for suggestions, I don't have enough credit to +1 useful answers.

Your solution seems slow because it iterates through the list (i.e. data) every time you make an update. A better approach would be using a dictionary, which takes O(1) as opposed to O(n) per update.
Here would be my solution using a dictionary:
file = open('input.txt', 'r')
#coordinates
c = {}
for line in file:
#items
(x, y, z) = (float(n) for n in line.split())
if (x, y) not in c:
c[(x, y)] = [z, 1]
elif c[(x, y)][0] > z:
c[(x, y)][0], c[(x, y)][1] = z, 1
elif c[(x, y)][0] == z:
c[(x, y)][1] += 1
for key in c:
print("{} {} {} {}".format(key[0], key[1], c[key][0], c[key][1]))

Why not change the last if to an elif ?
Like it is done now you will evaluate the z < data[i][2]: every iteration of the loop.
You could even just replace it with an else since you have already checked if z>data[i][2] and z == data[i][2] so the only remaining possibility is z < data[i][2]:
So following code will do the same and should be faster :
if z > data[i][2]:
continue
elif z==data[i][2]:
ctr = ctr + 1
data[i][3]=ctr
else:
data[i][2] = z
ctr = 1
data[i][3]=ctr

To do this in numpy use np.unique.
def count_unique(arr):
row_view=np.ascontiguousarray(a).view(np.dtype((np.void,a.dtype.itemsize * a.shape[1])))
ua, uind = np.unique(row_view,return_inverse=True)
unique_rows = ua.view(a.dtype).reshape(ua.shape + (-1,))
count=np.bincount(uind)
return np.hstack((unique_rows,count[:,None]))
First lets check for a small array:
a=np.random.rand(10,3)
a=np.around(a,0)
print a
[[ 0. 0. 0.]
[ 0. 1. 1.]
[ 0. 1. 0.]
[ 1. 0. 0.]
[ 0. 1. 1.]
[ 1. 1. 0.]
[ 1. 0. 1.]
[ 1. 0. 1.]
[ 1. 0. 0.]
[ 0. 0. 0.]]
print output
[[ 0. 0. 0. 2.]
[ 0. 1. 0. 1.]
[ 0. 1. 1. 2.]
[ 1. 0. 0. 2.]
[ 1. 0. 1. 2.]
[ 1. 1. 0. 1.]]
print np.sum(output[:,-1])
10
Looks good! Now lets check for a large array:
a=np.random.rand(3E7,3)
a=np.around(a,1)
output=count_unique(a)
print output.shape
(1331, 4) #Close as I can get to 600 unique elements.
print np.sum(output[:,-1])
30000000.0
Takes about 33 second on my machine and 3GB of memory, doing this all in memory for large arrays will likely be your bottleneck. For a reference #Joowani's solution took about 130 seconds, although this is a bit of an apple and oranges comparison as we start with a numpy array. Your milage may vary.
To read in the data as a numpy array I would view the question here, but it should look something like the following:
arr=np.genfromtxt("./input.txt", delimiter=" ")
Loading in that much data from a txt file I would really recommend using the pandas example in that link.

vectorize is indeterminate

I'm trying to vectorize a simple function in numpy and getting inconsistent behavior. I expect my code to return 0 for values < 0.5 and the unchanged value otherwise. Strangely, different runs of the script from the command line yield varying results: sometimes it works correctly, and sometimes I get all 0's. It doesn't matter which of the three lines I use for the case when d <= T. It does seem to be correlated with whether the first value to be returned is 0. Any ideas? Thanks.
import numpy as np
def my_func(d, T=0.5):
if d > T: return d
#if d <= T: return 0
else: return 0
#return 0
N = 4
A = np.random.uniform(size=N**2)
A.shape = (N,N)
print A
f = np.vectorize(my_func)
print f(A)
$ python x.py
[[ 0.86913815 0.96833127 0.54539153 0.46184594]
[ 0.46550903 0.24645558 0.26988519 0.0959257 ]
[ 0.73356391 0.69363161 0.57222389 0.98214089]
[ 0.15789303 0.06803493 0.01601389 0.04735725]]
[[ 0.86913815 0.96833127 0.54539153 0. ]
[ 0. 0. 0. 0. ]
[ 0.73356391 0.69363161 0.57222389 0.98214089]
[ 0. 0. 0. 0. ]]
$ python x.py
[[ 0.37127366 0.77935622 0.74392301 0.92626644]
[ 0.61639086 0.32584431 0.12345342 0.17392298]
[ 0.03679475 0.00536863 0.60936931 0.12761859]
[ 0.49091897 0.21261635 0.37063752 0.23578082]]
[[0 0 0 0]
[0 0 0 0]
[0 0 0 0]
[0 0 0 0]]

If this really is the problem you want to solve, then there's a much better solution:
A[A<=0.5] = 0.0
The problem with your code, however, is that if the condition passes, you are returning the integer 0, not the float 0.0. From the documentation:
The data type of the output of vectorized is determined by calling the function with the first element of the input. This can be avoided by specifying the otypes argument.
So when the very first entry is <0.5, it tries to create an integer, not float, array.
You should change return 0 to
return 0.0
Alternately, if you don't want to touch my_func, you can use
f = np.vectorize(my_func, otypes=[np.float])